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Mirrors > Home > ILE Home > Th. List > fneq2d | GIF version |
Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.) |
Ref | Expression |
---|---|
fneq2d.1 | ⊢ (φ → A = B) |
Ref | Expression |
---|---|
fneq2d | ⊢ (φ → (𝐹 Fn A ↔ 𝐹 Fn B)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | fneq2d.1 | . 2 ⊢ (φ → A = B) | |
2 | fneq2 4931 | . 2 ⊢ (A = B → (𝐹 Fn A ↔ 𝐹 Fn B)) | |
3 | 1, 2 | syl 14 | 1 ⊢ (φ → (𝐹 Fn A ↔ 𝐹 Fn B)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 98 = wceq 1242 Fn wfn 4840 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1333 ax-gen 1335 ax-4 1397 ax-17 1416 ax-ext 2019 |
This theorem depends on definitions: df-bi 110 df-cleq 2030 df-fn 4848 |
This theorem is referenced by: fneq12d 4934 |
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